tile at the same level ast. The link mapping may also be applied to element-tiles:L(ε)is the finer-link ofε, i.e., the set of potentially multiple overlapping resolving tiles at the same level asε;L∗(ε)is the coarser-link ofε, i.e., the set of overlapping resolving tiles at the next-coarser level fromε.
The evaluation-list ofIntegrateconsists of all red tiles, each one accompanied by a table of overlapping active functions.
We shall see that the nature of TCP is that an incremental change in the active setBleads to an incremental change in the tile coloring: this invites an incremental approach to coloring with consequent economy in refinement and integration.
Theorem 1 (activation) Suppose: (1)Sis consistent, (2)φˆ6∈ B, and (3)B=B+ ˆφ. Sis consistent iff three conditions hold:
1. E=E ∪ S( ˆφ)
In activatingφ, we might activate elements.ˆ
2. Bs(ε) =
Bs(ε) + ˆφ ε∈ S( ˆφ) update required Bs(ε) ε6∈ S( ˆφ) not affected
In activatingφ, we might update some native integration-tables.ˆ
3. Ba(ε) =
S
ε0∈D?(ε)Bs(ε0) Bs(ε) =∅ ∧ε∈ S( ˆφ) initialize Ba(ε) + ˆφ Bs(ε)6=∅ ∧ Bs(ε)6=∅ ∧ε∈S
ε0∈S( ˆφ)D(ε0) update Ba(ε) Bs(ε) =∅ ∨
Bs(ε)6=∅ ∧ε6∈S
ε0∈S( ˆφ)D(ε0)
not affected In activatingφ, we might update ancestral integration-tables.ˆ
Proof Please refer to this chapter’s Appendix for the formal proof.
4.3.2 Deactivation of Basis Function
Suppose we are in consistent stateS, and we deactivate some functionφ. What does the new stateˆ Slook like?
Theorem 2 (deactivation) Suppose: (1)Sis consistent, (2)φˆ∈ B, and (3)B=B −φ.ˆ Sis consistent iff three conditions hold:
1. E=E −n
ε∈ S( ˆφ)| Bs(ε) =∅:εo
In deactivatingφ, we might deactivate elements.ˆ
2. Bs(ε) =
Bs(ε)−φˆ ε∈ S( ˆφ) update required Bs(ε) ε6∈ S( ˆφ) not affected
In deactivatingφ, we might update some native integration-tables.ˆ
3. Ba(ε) =
∅ Bs(ε) =∅ ∧ε∈ S( ˆφ) clear Ba(ε)−φˆ Bs(ε)6=∅ ∧ε∈S
ε0∈S( ˆφ)D(ε0) update Ba(ε) Bs(ε)6=∅ ∧ε6∈S
ε0∈S( ˆφ)D(ε0)∨
Bs(ε) =∅ ∧ε6∈ S( ˆφ)
not affected In deactivatingφ, we might update ancestral integration-tables.ˆ
Proof Please refer to this chapter’s Appendix for the formal proof.
4.3.3 Refinement by Substitution
With theorems for activation and deactivation in place, we can easily prove theorems for compound opera- tions, such as substitution refinement.
Theorem 3 (substitution) An application of Deactivate composed with multiple applications of Activate, as shown below, is safe and effects a refinement ofφˆby substitution:
<S¯is consistent ∧ φˆ∈B¯> φˆactive, state is consistent
<{ϕ1, . . . , ϕN}=C( ˆφ)\B¯> give names to the inactive children ofφˆ S1←Activate( ¯S, ϕ1) activate first inactive child
. . . . . .
SN ←Activate(SN−1, ϕN) activateNth inactive child S ←Deactivate(SN,φ)ˆ deactivateφˆ
< Sis consistent> state is consistent
<φˆ6∈S.B ∧ C( ˆφ)⊂S.B> φˆwas replaced by its children
Proof sketch Compose the deactivation theorem withNapplications of the activation theorem.
4.3.4 Integration of Bilinear Forms
The specification of the evaluation list, (4.2), maps directly into an efficient algorithm, presented in Section 4.4.5.
4.3.5 Integration over Tiles
Lemma 1 (black tiles) Every ancestor element of a black element-tile is black.
Proof We prove that every elementt2, ancestor of black element-tilet, is black. Sincetis black, then it has an active descendant (by TCP1 ont). Any descendant oftis a descendant oft2, thust2 has an active descendant, andt2is black (by TCP1 ont2).
Lemma 2 (red tiles) Every descendant tile of a red tile is white.
Proof sketch Case 1: Red element tilet. Consider any particular element descendantt1. It is inactive (by TCP1 ont) thus it is not red (by TCP2 ont1). None of its descendants are active, sinceD(t1)⊂ D(t), thus it is not black (by TCP1 ont1). Therefore it is white. Consider any particular resolving-tile descendantt2. Its parent is not black (proof: chooset1to be the parent). Thereforet2is white (by TCP6 ont2).
Case 2: Red resolving tilet. Its child, the elementt3, is white (by TCP4 ont). Every descendant oft3is white (by the same argument as Case 1).
Lemma 3 (white tiles) The coarser-link of a white tile is not black.
Proof sketch Case 1: White element tilet. Assume its resolving-tile parentt1is black. Then the parent of t1is black (by TCP5 ont1). But thent1is red (by TCP4 ont1) which is a contradiction. Therefore the parent oftis not black.
Case 2: White resolving tilet. Assume its element-tile parentt2is black. Thentis either red (by TCP4 ont) or black (by TCP5 ont), which is a contradiction. Therefore the parent oftis not black.
Theorem 4 (tile coloring produces minimal valid partition) The red tiles, specified by the tile coloring problem, form a minimal partition of the domain that resolves every active element.
Proof sketch In Part I, we prove that the red tiles form a valid tiling. In part II, we show that the tiles are not excessively fine.
Part I The red tiles form a valid tiling, i.e., (1) the red tiles do not overlap, (2) every leaf element is a red tile, and (3) the red tiles cover the domain. Together (1) and (3) guarantee that the red tiles are a partition of the domain, and (2) guarantees that they resolve the finest active elements.
Assume two red tiles overlap, then one must be a descendant of the other. But by the Lemma 2 the descendants of a red tile are white. Therefore, (1) red tiles do not overlap.
By definition, a leaf element does not satisfy TCP1, therefore, by TCP2, (2) every leaf element is a red tile.
Pick any pointP on the domain. We show how to find the red tile that containsP. Choose the coarsest- level element tile containingP. By construction that element tile is red or black, since the active elements cover the domain. If it is red, QED; assume it is black. Traverse down the hierarchy of tiles as follows:
arriving at an element tile, proceed to the resolving tile containingP; arriving at resolving tile, proceed to its child tile. At every step of the traversal, examine the color of the tile. It must be black or red: it cannot be white by Lemma 3. If it is red, QED. If it is black, continue the traversal. Assume that the finest element is at levelq. Then there are no black tiles at levels≥q. Consequently the traversal must reach a red tile.
Therefore, (3) the red tiles cover the domain.
Part II The red tiles are not excessively-fine in the following sense: choose any element which has no active descendants. That element tile has no red descendants.
By construction the element is not black. If it is red then, by Lemma 2, QED. If it is white, then all of its descendants (element- and resolving-tiles) are white. To see this: observe that each of its element-descendants also has no active descendants hence is also white; consequently all its resolving-tile descendants are also white.